Question: Let $h(x)=\dfrac{e^x}{x-1}$ Where does $h$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=0$ (Choice B) B $x=1$ (Choice C) C $x=2$ (Choice D) D $h$ has no critical points.
Answer: A critical point of $h$ is a point in the domain of $h$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $h$, let's find its derivative. $\begin{aligned} h'(x)&=\dfrac{d}{dx}\left[ \dfrac{e^x}{x-1} \right] \\\\ &=\dfrac{\dfrac{d}{dx}[e^x](x-1)-(e^x)\dfrac{d}{dx}[x-1]}{(x-1)^2} \\\\ &=\dfrac{e^x(x-1)-e^x(1)}{(x-1)^2} \\\\ &=\dfrac{e^x(x-2)}{(x-1)^2} \end{aligned}$ Now let's look for $x$ -values where $h'$ is zero or undefined. $\dfrac{e^x(x-2)}{(x-1)^2}=0$ at $x=2$. $\dfrac{e^x(x-2)}{(x-1)^2}$ is undefined at $x=1$. However, $h(x)=\dfrac{e^x}{x-1}$ is also undefined at $x=1$ so it isn't a critical point. In conclusion, this is the only $x$ -value where $h$ has a critical point: $x=2$